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a year ago

Physics Atoms Atoms

A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of:

(Given, Planck's constant = 6.6 * 10^-³⁴ Js).

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Vishal Baghel

Contributor-Level 10

Given absorb energy = 10.2 eV

$E_{f} = \frac{- 13.6}{n^{2}} = - 3.4$

$n^{2} = 4$

n = 2

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a year ago

Physics Atoms Atoms

The momentum of an electron revolving in nth orbit is given by:

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Vishal Baghel

Contributor-Level 10

mvr = $\frac{n h}{2 π}$ $\frac{}{}$

So momentum (mv) = $\frac{n h}{2 π r}$

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Physics Atoms Class 12th

Hydrogen atom from exited state comes to the ground state by emitting a photon of wavelength $λ .$ The value of principal quantum number 'n' of the excited state will be

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Payal Gupta

Contributor-Level 10

According to definition of wave number, we can write

$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow 1 - \frac{1}{n^{2}} = \frac{1}{λ R}$

$\Rightarrow \frac{1}{n^{2}} = 1 - \frac{1}{λ R} = \frac{λ R - 1}{λ R} \Rightarrow n = \sqrt[]{\frac{λ R}{λ R - 1}}$

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Physics Atoms Class 12th

$\frac{x}{x + 4}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its

(i) third permitted energy level to the second level and

(ii) the highest permitted energy level to the second permitted level.

The value of x will be ________.

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Payal Gupta

Contributor-Level 10

According to definition of atomic energy, we can write

$E_{1} = E_{0} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 E_{0}}{36}, a n d E_{2} = E_{0} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$

$\frac{E_{1}}{E_{2}} = \frac{5}{9} \Rightarrow \frac{x}{x + 4} = \frac{5}{5 + 4} \Rightarrow x = 5$

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Physics Atoms Class 12th

In Bohr's atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level

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Payal Gupta

Contributor-Level 10

Based on theory

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a year ago

Physics Atoms Atoms

Choose the correct option from the following options given below :

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Vishal Baghel

Contributor-Level 10

According to Rutherford, e- revolves around in nucleus in circular orbit. Thus e- is always accelerating (centripetal acceleration). An accelerating change emits EM radiation and thus e- should loose energy and finally should collapse in the nucleus.

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